S.No.  Object-Distance u( {\cm})  Image-Distancev( {\cm})  1  -90  +18  2  -60  +20  3  -30  +30  4  -20  +60  5  -18  +90  6  -10  +100" >  S.No.  Object-Distance u( {\cm})  Image-Distancev( {\cm})  1  -90  +18  2  -60  +20  3  -30  +30  4  -20  +60  5  -18  +90  6  -10  +100">


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CBSE Class 10 Science 2017 Delhi Term II Set 1 Paper

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Question : 22
Total: 36
Analyse the following observation table showing variation of image distance (v) with object distance (u) in case of a convex lens and answer the questions that follow, without doing any calculations:div class="hscrollenable">
 S.No.  Object-Distance
u(cm)		  Image-Distance
v(cm)
 1  −90  +18
 2  −60  +20
 3  −30  +30
 4  −20  +60
 5  −18  +90
 6  −10  +100
Solution:  
(a) From S. No. 3, we can say that the radius of curvature is 30‌cm because when an object is placed at the centre of curvature of a convex lens, its image is formed on the other side of the lens at the same distance from the lens. And, we also know that focal length is half of the radius of curvature. Therefore, focal length of the lens is +15‌cm.
(b) S. No. 6 is not correct as the object distance is between focus and optical centre, and in such cases, the image formed is always virtual. But in this case, a real image is formed as the image distance is positive. Therefore, v should be negative.
(c) Approximate value of magnification for object distance −20‌cm and image distance + 60‌cm is -3 .
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